Let
be the function given by
 = |\theta|)
for
![\theta\in [ -\pi , \pi ]](http://latex.codecogs.com/png.latex?\theta\in [ -\pi , \pi ])
.
a) Calculate the fourier coefficients of
and show that...
 = \frac{(-1)^n - 1}{\pi n^2})
for
 = \frac{\pi}{2})
for
.
done...
b) What is the Fourier series of
in terms of sines and cosines?
My answer,
^n - 1}{\pi n^2} \cos(n\theta))
.
For even n's the summands of this series are zero, so you can take only odd n's in the sum and since you have a square in the denominator this is the same as twice the sum over the odd natural numbers (taking 
): ^n - 1}{\pi n^2} \cos(n\cdot 0)=\frac{\pi}{2}+\frac{2}{\pi}\sum^\infty_{k=0}\fr ac{-2}{(2k-1)^2})
^2})
. Now equal the above to zero and you'll get what you want... Tonio
c) Taking
, prove that
^2} = \frac{\pi^2}{8})
and
Now this, I suspect, should have just been a simple application of Parsevals theorem but I'm going wrong somewhere.
Parsevals Theorem. |^2 d \theta \right)^{\frac{1}{2}} = \left ( \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2\right)^{\frac{1}{2}})
But with
the left hand side is 0.
Thus we get,
|^2\right)^{\frac{1}{2}})
=
^2 \right)^{\frac{1}{2}} = \frac{\pi}{2})
So clearly I'm doing something wrong... Most likely in that last 2 steps...
Originally Posted by Deadstar
Let
LinkBack URL
About LinkBacks